Consider the Bott projection (described in Exercise 5.I of Wegge-Olsen's book $K$-theory and $C^*$-algebras) given by $b(z)=\frac{1}{1+|z|^2}\begin{pmatrix} 1 & \bar{z} \\ z & |z|^2 \end{pmatrix}$. Regarded as an element in $C_0(\mathbb{R}^2,M_2(\mathbb{C}))$ via $z=x+iy$, $b$ represents a class in $K_0(C_0(\mathbb{R}^2))$. Under the suspension isomorphism $K_1(A)\cong K_0(SA)$ for any $C^*$-algebra $A$, which tells us in particular $K_0(C_0(\mathbb{R}^2))\cong K_1(C(S^1))$, what is the class corresponding to $b$ in $K_1(C(S^1))$? I think it should be the class of the function $z\mapsto\bar{z}$ but I cannot prove it.


I don't know if it will be helpful for you, but I'll try:

I'm not sure how you see the isomorphism between $K_0(C_0(\Bbb{R}^2))$ and $K_1(C(S^1))$ by the suspension isomorphism.
However, consider the short-exact sequence $0\to C_0(\Bbb{R}^2)\to C(\Bbb{D})\to C(S^1)\to 0$ where the maps are inclusion $\phi$ and restriction $\psi$ and you identify $\Bbb{R}^2$ with $\Bbb{D}-S^1$.
It induces a six-term exact sequence (couldn't write it here, but you know what I mean). We have: $K_1(C(S^1))\cong K_0(C(S^1)) \cong K_0(C(\Bbb{D})\cong \Bbb{Z}$ ; $K_1(C(\Bbb{D}))=0$ ;and the generator of $K_0(C(S^1))$ is $[1]_0$ . Conclude that $K_0(\psi)$ is an isomorphism (because it maps generator to generator) and thus $\delta: K_1(C(S^1)) \to K_0(C_0(\Bbb{R}^2))$ , the index map, is an isomorphism!
Now, you can check what $\delta$ does to the generator of $K_1(C(S^1))$, i.e. $[f(z)=z]_1$, to determine $\delta$ in general and to find the generator of $K_0(C_0(\Bbb{R}^2))$. After some calculations I have come to a conclusion that $\delta([z]_1)=\begin{bmatrix}|z|^2&z(1-|z|^2)^{1/2}\\ \bar{z}(1-|z|^2)^{1/2}&1-|z|^2\end{bmatrix}_0-\begin{bmatrix}1&0\\0&0\end{bmatrix}_0 \in K_0(C_0(\Bbb{R}^2))$
My writing is a little bit confusing, I mean the equivalence classes of the matrices.
$[\bar{z}]_1=-[z]_1$ in $K_1(C(S^1))$, so you can easily see which element of $K_0(C_0(\Bbb{R}^2))$ mapped to $[\bar{z}]_1$.
See if it helps, I'm not familiar yet with the Bott projection.


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